NIMCET 2014 — Mathematics PYQ

Concept

• Logarithms: If $a^x=b$, then we say that ${\log }_{a}b=x$.

• $\log (ab)=\log a+\log b$.

Calculation

Let's say that $\ln 2=a$ and $\ln 3=b$.

Given equation: $(2x{)}^{\ln 2}=(3y{)}^{\ln 3}$

$\Rightarrow (2x{)}^a=(3y{)}^b$

Taking the natural log of both sides, we get:

$\Rightarrow a(\ln 2x)=b(\ln 3y)$

$\Rightarrow a(\ln 2+\ln x)=b(\ln 3+\ln y)$

$\Rightarrow a(a+\ln x)=b(b+\ln y)$

$\Rightarrow a^2+a\ln x=b^2+b\ln y$

$\Rightarrow a\ln x-b\ln y=b^2-a^2...(1)$

Given second equation: $3^{\ln x}=2^{\ln y}$

Taking the natural log of both sides, we get:

$\Rightarrow (\ln x)(\ln 3)=(\ln y)(\ln 2)$

$\Rightarrow \ln y=\frac{b}{a}(\ln x)...(2)$

Substituting equation (2) in equation (1), we get:

$\Rightarrow a\ln x-b\times \frac{b}{a}(\ln x)=b^2-a^2$

$\Rightarrow (\ln x)\left(\frac{a^2-b^2}{a}\right)=b^2-a^2$

$\Rightarrow \ln x=-a=-\ln 2$

$\Rightarrow \ln x=\ln 2^{-1}$

$\Rightarrow x=2^{-1}=\frac{1}{2}$

Therefore, the solution $x_0$ is $\frac{1}{2}$.

Correct Option: 3